Re: Proof that A^n = 0 if A is nilpotent
- From: Michael Press <rubrum@xxxxxxxxxxx>
- Date: Thu, 13 Dec 2007 02:15:35 GMT
In article
<abe5b6f5-eee7-48b9-939d-aae7a846dbe7@xxxxxxxxxxxxxxxxx
roups.com>,
vip.maverick@xxxxxxxxx wrote:
Hi ,
I would like to request some help if anyone has ideas about how to
prove that if A is nilpotent matrix then A^n = 0?
Take the dimension of A to be n.
Suppose s is the least s such that A^s = 0
Take u =/= 0 in R^n.
There is a least integer k <= n such that
A^0.u, A^1.u, A^2.u, ..., A^k.u are linearly dependent.
That is to say there is a polynomial P of degree k <= n
such that P(A) = 0. Therefore P(X) divides X^s, and P(X) = X^k.
Therefore s <= n, and A^n = 0.
--
Michael Press
.
- References:
- Proof that A^n = 0 if A is nilpotent
- From: vip . maverick
- Proof that A^n = 0 if A is nilpotent
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